# Interesting Puzzle

• Nov 6th 2009, 08:06 PM
everydayangel
Interesting Puzzle
I liked this one... Had it in a math competition...

Replace each letter of POPLAR with a digit 0 through 9 (equal letters replaced by equal digits, different letters replaced by different digits). If the resulting number is the largest such number divisible by 55, find P + O + P + L + A + R

I had a multiple choice to choose from... see attached for multiple choice if you must(Tongueout)...

Happy Trails! ;-)

Everydayangel
(Angel)
• Nov 6th 2009, 10:39 PM
Gusbob
Were you allowed a calculator? If so, I have an excellent brute force method that will take seconds. If not, my method will still work but will take minute or two longer to manually calculate.

Since 55 is relatively small number, it has many multiples within any range of 1000. So it really doesn't matter what the first three digits are in terms of divisibility. So I pick the highest possible 3: 989xxx. Consequently, I intend to find the lowest 7 digit multiple of 55 and work backwards from there. I find that the factor that yields this lowest 7 digit number is 1 000 000 / 55 = 18181.81818...

So the factor multiplied with 55 to form POPLAR is probably a few hundred integers under that. (because there is a minimum difference between 1000000 and 989xxx of 10000, I reuse the above calculation 10 000 / 55 = 181.818...)

So the factor is maximum 18181 - 181 = 18000.

18000 x 55 = 990000
17999 x 55 = 990000 - 55 = 989945. But P and L have the same values, so this is not the multiple
17998 x 55 = 989945 - 55 = 989890. But P and A have the same values, so this is not the multiple
And keep on going down the list and come to:
17995 x 55 = 989725.

So P + O + P + L + A + R = 9 + 8 + 9 + 7 + 2 + 5 = 40

I know this method is very crude. I would like to know if there is a better way of doing this.
• Nov 7th 2009, 07:21 AM
Wilmer
Highest possible is 989765 (digits descending order; 9 repeated)

989765 / 55 = 17995.727....

17995 * 55 = 989725
• Nov 7th 2009, 11:47 PM
everydayangel
Wow! Good job.
Technically, I think that your way was easier. I found the highest power for 55 that would give me the highest six digit # without going over 989xxx with a calculator (took me a bit) then worked up to the answer. Great job! I've got more too from that competition. Since they don't reuse the problems they let me keep a copy of each round of tests. Cool huh? Yeah I'm a dork! (Rofl)

anyhow- good job. I'll have to remember what you did.
Ang(Angel)
• Nov 8th 2009, 03:32 PM
Soroban
Hello, everydayangel!

Quote:

Replace each letter of POPLAR with a digit 0 through 9
(equal letters replaced by equal digits, different letters replaced by different digits).
If the resulting number is the largest such number divisible by 55, find P + O + P + L + A + R

The number is divisible by 5 and 11.

To be divisible by 5, it must end in 0 or 5.

To be divisible by 11, (sum of 1st, 3rd. 5th, etc. digits) - (sum of 2nd, 4th, 6th, etc. digits)
. . must be a multiple of 11.

Seeking the largest number, let $\displaystyle P = 9, O = 8$

We have: .$\displaystyle 989LAR$
. . where: .$\displaystyle R = 0\,\text{ or }\,5$ [1]
. . and: .$\displaystyle (9+9+A) - (8 + L + R) \:=\:11k\,\text{ for some integer }k.$ [2]

From [2], we have: .$\displaystyle 10 - L + A - R \:=\:11k$

The largest number would have $\displaystyle L = 7\!:\;\;3 + A - R \:=\:11k$

If $\displaystyle R = 0$, then $\displaystyle A = 8$, which is not allowed (since $\displaystyle O = 8$).

Hence, $\displaystyle R = 5\,\text{ and }\,A = 2$

Therefore: .$\displaystyle POPLAR \:=\:989725\:\text{ and }\:P+O+P+L+A+R \:=\:40$